A) 70
B) 50
C) 110
D) 55
Correct Answer: D
Solution :
\[x+\frac{1}{x}=5\] \[\Rightarrow \]\[{{x}^{2}}-5x+1=0\Rightarrow {{x}^{2}}-3x+1=2x\] \[\therefore \]\[\frac{{{x}^{4}}+\frac{1}{{{x}^{2}}}}{{{x}^{2}}-3x+1}=\frac{1}{2}\left( \frac{{{x}^{2}}+\frac{1}{{{x}^{2}}}}{x} \right)=\frac{1}{2}\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)\] \[=\frac{1}{2}\left[ {{\left( x+\frac{1}{x} \right)}^{3}}-3\left( x+\frac{1}{x} \right) \right]\] \[=\frac{1}{2}(125-3\times 5)=\frac{1}{2}\times 110=55\]
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